Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main20265 April 2026Morning ShiftMathematicsLimitsActual

The product of all possible values of , for which _ x 0 ( 1 - ( x) (( +1)x) (( +2)x) ^2(( +1)x) ) = 2 , is:

Options

  1. A-2
  2. B1
  3. C-1
  4. D5 4

Correct answer

C. -1

Step-by-step solution

Using the standard expansions for small x , 1 - ^2 2 and . The given limit can be written as: _ x 0 1 - (1 - ^2 x^2 2 ) (1 - ( +1)^2 x^2 2 ) (1 - ( +2)^2 x^2 2 ) (( +1)x)^2 Neglecting higher powers of x , the numerator simplifies to: 1 - (1 - x^2 2 ( ^2 + ( +1)^2 + ( +2)^2 ) ) = x^2 2 ( ^2 + ( +1)^2 + ( +2)^2 ) Substituting this back into the limit: _ x 0 x^2 2 ( ^2 + ( +1)^2 + ( +2)^2 ) ( +1)^2 x^2 = ^2 + ( +1)^2 + ( +2)^2 2( +1)^2 We are given that this limit is equal to 2 . Therefore: ^2 + ( +1)^2 + ( +2)^2 2( +

Practice Limits on Quantrex Academy →

More from Limits

Let _ x 2 ( (x-2))(rx^2 + (p-2)x - 2p) (x-2)^2 = 5 for some r, p R . If the set of all possible values of q , such that the roots of the equation rx^2 - px + q = 0 lie in (0, 2) , 2026The value of _ x 0 ( x^2 ^2 x x^2 - ^2 x ) is: 2026Let f(x) = _ y 0 (1 - (xy)) (xy) y^3 . Then the number of solutions of the equation f(x) = x , x R is : 2026If _ x 2 (x^3 - 5x^2 + ax + b) ( x-1 - 1) _e(x-1) = m , then a + b + m is equal to : 2026The value of _ x 0 _ e ( (e x) (e² x ) (e¹⁰ x ) ) e²-e^ 2 x is equal to 2026If _ x 0 e ^ ( a -1) x +2 ~b x+( c -2) e ^ -x x x- _ e (1+x) =2 , then a ²+ b ²+ c ² is equal to : 2026Let [ ] denote the greatest integer function and f(x)= _ n 1 n ³ _ k =1 ^ n [ k ² 3^ x ] . Then 12 _ j =1 ^ f( j ) is equal to _ _ _ _ . 2026Let f: R (0, ) be a twice differentiable function such that f(3)=18, f^ (3)=0 and f^ (3)=4 . Then _ x 1 ( _ e ( f(2+x) f(3) )^ 18 (x-1)² ) is equal to : 2026 Full Limits list All JEE Main PYQs